$(r_1 − r) (r_2 − r)(r_3 − r) = 4 Rr^ 2 $

Question

We have to prove $(r_1− r) (r_2− r)(r_3− r) = 4 Rr^2$

I know the following formula , But I could not understand how to use them

Answer 1

We know $$r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$$ Thus, $$r_1-r = 4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} -4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = 4R\sin ^2 \frac{A}{2}$$ Similarly it can be shown that: $$r_2-r = 4R\sin ^2 \frac{B}{2} \text{ and that } r_3-r = 4R\sin ^2\frac{C}{2}$$ Thus, $$(r_1-r)(r_2-r)(r_3-r) = 64R^3\sin^2\frac{A}{2}\sin ^2\frac{B}{2}\sin ^2\frac{C}{2} = 4R(4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})^2 = 4Rr^2$$ Hope it helps.

Answer 2

We have \begin{align} (r_1-r)(r_2-r)(r_3-r)&= \left(\frac{\Delta}{s-a} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s-b} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s-c}-\frac{\Delta}{s}\right)\\ &={\Delta}^3 \left(\frac{ abc}{s^3(s-a)(s-b)(s-c)}\right)\\ &=\frac{abc{\Delta}^2}{s^2\Delta}\\ &=4R{\left(\frac{\Delta}{s}\right)}^2\\ &=4Rr^2. \end{align}